What trigonometric identity is used to simplify the initial expression?

The expression is simplified using the angle addition/subtraction formulas, specifically the expansion for sin(A + B) or cos(A - B) after rearranging terms.

How do we determine sin θ and cos θ from cot θ in the second quadrant?

Since π/2 < θ < π, sin θ is positive and cos θ is negative. Using cot θ = -1/2√2, we find sin θ = 2√2/3 and cos θ = -1/3.

What is the value of θ if cos θ = -1/3?

While θ is not a standard angle, we can express the final result in terms of θ or find its relation to π/2 and π for reduction.

How does the expression simplify to sin(8θ + 15θ/2) + cos(8θ + 15θ/2)?

By distributing the terms and grouping them, the expression fits the pattern sin A cos B + cos A sin B + cos A cos B - sin A sin B, which is sin(A+B) + cos(A+B).

What is the final angle inside the trig functions?

The combined angle is (8θ + 15θ/2) which equals 31θ/2.

Why is the answer negative?

The sign depends on the quadrant of the resulting angle 31θ/2. Based on the range of θ, this angle falls into a quadrant where the combined sine and cosine values result in (1-√2)/√3.

Can this be solved using complex numbers (Euler's formula)?

Yes, representing the terms as real/imaginary parts of e^(iθ) can streamline the simplification of products of sines and cosines.

What is the common mistake in this problem?

The most common error is failing to account for the second quadrant signs for sin and cos when calculating their values from cot θ.

How do we handle the half-angle 15θ/2?

The half-angle is best handled by keeping it in the addition formula until the final step where the values of sin θ and cos θ are substituted.

Is this considered a difficult problem?

It is considered medium-to-hard due to the algebraic manipulation required before the trigonometric values can be substituted.

Let π/2 < θ < π and cot θ = -1/2√2. Then the value of sin(15θ/2)(cos 8θ + sin 8θ) + cos(15θ/2)(cos 8θ

Q: How do we determine sin θ and cos θ from cot θ in the second quadrant?

Since π/2 < θ < π, sin θ is positive and cos θ is negative. Using cot θ = -1/2√2, we find sin θ = 2√2/3 and cos θ = -1/3.

Q: What is the value of θ if cos θ = -1/3?

While θ is not a standard angle, we can express the final result in terms of θ or find its relation to π/2 and π for reduction.

Q: How does the expression simplify to sin(8θ + 15θ/2) + cos(8θ + 15θ/2)?

By distributing the terms and grouping them, the expression fits the pattern sin A cos B + cos A sin B + cos A cos B - sin A sin B, which is sin(A+B) + cos(A+B).

Q: What is the final angle inside the trig functions?

The combined angle is (8θ + 15θ/2) which equals 31θ/2.

Q: Why is the answer negative?

The sign depends on the quadrant of the resulting angle 31θ/2. Based on the range of θ, this angle falls into a quadrant where the combined sine and cosine values result in (1-√2)/√3.

Q: Can this be solved using complex numbers (Euler's formula)?

Yes, representing the terms as real/imaginary parts of e^(iθ) can streamline the simplification of products of sines and cosines.

Q: What is the common mistake in this problem?

The most common error is failing to account for the second quadrant signs for sin and cos when calculating their values from cot θ.

Q: How do we handle the half-angle 15θ/2?

The half-angle is best handled by keeping it in the addition formula until the final step where the values of sin θ and cos θ are substituted.

Q: Is this considered a difficult problem?

It is considered medium-to-hard due to the algebraic manipulation required before the trigonometric values can be substituted.

Let π/2 < θ < π and cot θ = -1/2√2. Then the value of sin(15θ/2)(cos 8θ + sin 8θ) + cos(15θ/2)(cos 8θ - sin 8θ) is equal to : | JEE Main Mathematics

JEERANKERS

Math Trigonometry

Q MCQ Angle Reduction

Let $\frac{\pi}{2} < \theta < \pi$ and $\cot \theta = -\frac{1}{2\sqrt{2}}$. Then the value of $\sin \left( \frac{15\theta}{2} \right)(\cos 8\theta + \sin 8\theta) + \cos \left( \frac{15\theta}{2} \right)(\cos 8\theta - \sin 8\theta)$ is equal to :

A) $\frac{\sqrt{2}-1}{\sqrt{3}}$ B) $\frac{\sqrt{2}}{\sqrt{3}}$ C) $\frac{1-\sqrt{2}}{\sqrt{3}}$ D) $-\frac{\sqrt{2}}{\sqrt{3}}$

✅ Correct Answer

C) (1-√2)/√3

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Solution Steps

Expand and Regroup the Expression

The given expression is $E = \sin(\frac{15\theta}{2})\cos 8\theta + \sin(\frac{15\theta}{2})\sin 8\theta + \cos(\frac{15\theta}{2})\cos 8\theta - \cos(\frac{15\theta}{2})\sin 8\theta$.

Regrouping the terms:

$E = [\cos(\frac{15\theta}{2})\cos 8\theta + \sin(\frac{15\theta}{2})\sin 8\theta] + [\sin(\frac{15\theta}{2})\cos 8\theta - \cos(\frac{15\theta}{2})\sin 8\theta]$.

Apply Compound Angle Identities

Recall: $\cos(A-B) = \cos A \cos B + \sin A \sin B$ and $\sin(A-B) = \sin A \cos B - \cos A \sin B$.

Applying these to our grouped terms:

$E = \cos(8\theta - \frac{15\theta}{2}) - \sin(8\theta - \frac{15\theta}{2})$.

$E = \cos(\frac{\theta}{2}) - \sin(\frac{\theta}{2})$.

Determine sin θ and cos θ from cot θ

Given $\cot \theta = -\frac{1}{2\sqrt{2}}$ in the 2nd quadrant ($\frac{\pi}{2} < \theta < \pi$).

$\csc^2 \theta = 1 + \cot^2 \theta = 1 + \frac{1}{8} = \frac{9}{8}$.

$\sin \theta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$ (positive in 2nd quadrant).

$\cos \theta = \cot \theta \cdot \sin \theta = (-\frac{1}{2\sqrt{2}})(\frac{2\sqrt{2}}{3}) = -\frac{1}{3}$.

Find sin(θ/2) and cos(θ/2)

Since $\frac{\pi}{4} < \frac{\theta}{2} < \frac{\pi}{2}$, both $\sin(\frac{\theta}{2})$ and $\cos(\frac{\theta}{2})$ are positive.

$\cos(\frac{\theta}{2}) = \sqrt{\frac{1+\cos\theta}{2}} = \sqrt{\frac{1-1/3}{2}} = \sqrt{\frac{2/3}{2}} = \frac{1}{\sqrt{3}}$.

$\sin(\frac{\theta}{2}) = \sqrt{\frac{1-\cos\theta}{2}} = \sqrt{\frac{1+1/3}{2}} = \sqrt{\frac{4/3}{2}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}$.

Calculate Final Value

Substitute the half-angle values back into the expression for $E$:

$E = \cos(\frac{\theta}{2}) - \sin(\frac{\theta}{2}) = \frac{1}{\sqrt{3}} - \frac{\sqrt{2}}{\sqrt{3}}$.

$E = \frac{1 - \sqrt{2}}{\sqrt{3}}$

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Theory

1. Compound Angle Identities

Compound angle identities are fundamental formulas that relate the trigonometric functions of sums or differences of angles to the functions of the individual angles. The core identities used here are $\cos(A-B) = \cos A \cos B + \sin A \sin B$ and $\sin(A-B) = \sin A \cos B - \cos A \sin B$. These formulas allow us to collapse complex algebraic sums into single trigonometric terms, which is a key strategy in solving JEE trigonometry problems where the angles appear in linear combinations like $8\theta$ and $15\theta/2$.

2. Half-Angle Formulas

Half-angle formulas express the trigonometric functions of $\theta/2$ in terms of $\cos \theta$. They are derived from the double-angle formula for cosine: $\cos \theta = 2\cos^2(\theta/2) - 1 = 1 - 2\sin^2(\theta/2)$. By rearranging these, we get $\cos(\theta/2) = \pm \sqrt{(1+\cos\theta)/2}$ and $\sin(\theta/2) = \pm \sqrt{(1-\cos\theta)/2}$. The choice of sign depends entirely on the quadrant in which $\theta/2$ lies. In this problem, because $\pi/2 < \theta < \pi$, the half-angle $\theta/2$ is in the first quadrant, making both results positive.

3. Quadrant Rules (ASTC Rule)

The ASTC rule (All-Silver-Tea-Cups) determines the signs of trigonometric functions in the four quadrants. In the first quadrant (0 to $\pi/2$), all functions are positive. In the second quadrant ($\pi/2$ to $\pi$), only Sine and Cosecant are positive. In the third, Tangent and Cotangent are positive, and in the fourth, Cosine and Secant are positive. This problem requires moving between quadrants for $\theta$ (2nd quadrant) and its half-angle $\theta/2$ (1st quadrant) to assign the correct numerical signs to the calculated ratios.

4. Algebraic Trigonometry Manipulation

Often in competitive exams, the goal is not to substitute values immediately but to simplify the expression algebraically first. The provided expression was a linear combination of products. Distributing the terms allowed for regrouping into standard identity forms. This "look-before-you-leap" approach prevents tedious calculations with irrational numbers like $\sqrt{2}/3$ and $-1/3$ until the very end, reducing the risk of arithmetic errors during the simplification process.

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FAQs

Why did $\cos(8\theta - 15\theta/2)$ simplify to $\cos(\theta/2)$?

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Because $8\theta$ is equal to $16\theta/2$. Therefore, $16\theta/2 - 15\theta/2 = 1\theta/2$. The subtraction of coefficients makes the simplification straightforward.

What if $\theta$ was in the 4th quadrant?

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If $\theta$ was in the 4th quadrant, $\cos \theta$ would be positive and $\sin \theta$ would be negative. Additionally, $\theta/2$ would lie in the 2nd quadrant, affecting the signs of the half-angle results.

Is there a shortcut for $\cos A - \sin A$?

⌄

Yes, it can be written as $\sqrt{2} \cos(A + \pi/4)$ or $\sqrt{2} \sin(\pi/4 - A)$. However, direct substitution is usually faster once the half-angle values are known.

How do we know $\sin \theta$ is $2\sqrt{2}/3$?

⌄

Using the identity $\sin^2\theta = 1/(1+\cot^2\theta) = 1/(1+1/8) = 8/9$. Taking the square root gives $2\sqrt{2}/3$. It is positive because $\theta$ is in the 2nd quadrant.

Can we find $\theta$ exactly?

⌄

No, $\cos \theta = -1/3$ is not a standard angle value. We must use trigonometric ratios rather than the angle value itself to solve this problem.

What is the significance of the $15\theta/2$ and $8\theta$ angles?

⌄

They are chosen such that their sum or difference results in a simpler angle (like $\theta/2$ or $31\theta/2$). In this specific problem, their difference $\theta/2$ leads directly to the answer.

Does $\sin(-\theta) = -\sin \theta$ apply here?

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Yes, sine is an odd function. However, in our simplification $8\theta - 15\theta/2$, the result was positive $\theta/2$, so the odd function property wasn't needed.

Why is option B incorrect?

⌄

Option B is $\sqrt{2}/\sqrt{3}$, which is just the value of $\sin(\theta/2)$. The expression requires $\cos(\theta/2) - \sin(\theta/2)$, which includes the $1/\sqrt{3}$ term.

How much time should this take in a JEE exam?

⌄

Ideally, this should be solved in 2.5 to 3 minutes. The identification of the identity is the most time-saving step.

Is there any other identity for $\cos(x) - \sin(x)$?

⌄

One useful identity is $(\cos x - \sin x)^2 = 1 - \sin 2x$. This can be used if you know $\sin \theta$, but you'd still need to take a square root and check the sign.

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